1. Triangle ABC has side lengths AB=9, AC=10, and BC=17. Let X be the intersection of the angle bisector of angle A with side \(\overline{BC}\), and let Y be the foot of the perpendicular from X to side \(\overline{AC}\). Compute the length of \(\overline{XY} \)

2. \(\overline{BC}\) is a chord of a circle with center O and area \(48\pi \). Point A is inside \(\triangle BCO\) such that \(\triangle ABC\) is equilateral and A is the circumcenter of \(\triangle BCO\). What is the area of triangle ABC?

Thank you!!!

yasbib555 Apr 6, 2019

#1**0 **

First one....see the image below :

Let A = (0,0)

Let C = (10,0)

And we can find B, thusly

Construct a circle with radius 9 centered at the origin

The equation of this circle will be x^2 + y^2 = 81 (1)

And construct a circle wilth radius 17 centered at (10, 0)

The equation of this circle will be (x - 10)^2 + y^2 = 289 (2)

We can find the coordinates for point B, thusly :

Subtract (1) from (2) and we have that

(x-10)^2 - x^2 = 208 simplify

x^2 - 20x + 100 - x^2 = 208

-20x = 108

x = -108/20 = -5.4 = x coordinate of B

And taking the positive value for y, we have that

(-5.4)^2 + y^2 = 81

y^2 = 81 - 29.16

y = 7.2 = y coordinate of B

So....B = (-5.4, 7.2)

And AX is the bisector of angle A

So....we have the following relationship

AC/AB = XC /XB

10/9 = XC / XB

Then BC has 19 equal parts and XC is 10 of them.....so XC = (10/19)*17 = 170/19

So....using a similar triangle idea.....

Distance from B to the x axis / BC = XY / XC

7.2 / 17 = XY / (170/19)

(170/19) * 7.2 /17 = XY = (170/17)(7.2 / 19) = 10 * 7.2 / 19 = 72 / 19

CPhill Apr 6, 2019

#2**0 **

Second one.....see the following image....

Let O = (0,0)

Since the area of circle O = 48pi....then the radius = √48 units

So...the equation of the circle is x^2 + y^2 = 48

Since ABC is equilateral....then angle BAC = 60°

Then the excluded angle to BAC = 300°

So, by symmetry, angle AOC = 150°

And since A is the circumcenter of triangle BOC, then AO = AC = the side of equilateral triangle ABC = s

And we can find s^2 using the Law of Cosines, thusly :

OC^2 = 2s^2 - 2s^2 * ( cos 150)

48 = 2s^2 + 2s^2* (√3/2)

48 = 2s^2 + s^2√3

48 = (2 + √3)s^2

48 / [ 2 + √3 ) = s^2 = 48 [ 2 - √3] = 96 - 48√3

So....the area of equilateral triangle ABC =

(1/2) s^2 sin (BAC) =

(1/2)s^2 sin(60°) =

(√3) [ 96 - 48√3 ] / 4 =

(√3) [ 24 - 12√3] =

24√3 - 36 units^2

CPhill Apr 7, 2019